How do you find distance without knowing the time?

The trouble with x = v₀t + ½at²

There is nothing wrong with it. It just needs t, and plenty of questions never give you the time. You could find t first from v = v₀ + at and substitute. That's two steps, and every extra step is another chance to drop a sign.

NCERT gives the shortcut as equation (2.8):

v² = v₀² + 2ax

One caution about x. Here it is the displacement from the starting point (x = 0), not the distance travelled. For a car that never turns back, the two match. For a ball thrown up and caught again, they don't.

Where the equations come from

Section 2.4 builds them from one picture, Fig. 2.5. With constant acceleration, the velocity–time graph is a straight sloping line that starts at v₀ and reaches v after time t.

The area under that line is the displacement. Split it into two pieces:

• a rectangle (OACD in the figure) of height v₀ and width t, so area v₀t
• a triangle (ABC) sitting on top, with base t and height v − v₀ = at, so area ½at²

Add them and you get x = v₀t + ½at². Read the same shape as a trapezium and you get x = (v + v₀)t/2, which is just average velocity times time. Now remove t using v = v₀ + at, and v² = v₀² + 2ax falls out.

Strictly, area counts as displacement with a sign: area below the time axis is negative. In Fig. 2.5 the velocity stays positive, so all of it adds.

One motion, solved three ways

A cyclist moving at 5 m s⁻¹ speeds up at 2 m s⁻² for 3 s. After 3 s the velocity is 5 + 2 × 3 = 11 m s⁻¹.

• x = v₀t + ½at² = 15 + 9 = 24 m
• x = average velocity × t = 8 × 3 = 24 m
• x = (v² − v₀²) ÷ 2a = (121 − 25) ÷ 4 = 24 m

Same answer every time. In an exam that's a free check: if two methods disagree, one of them has a slip.

A tip for picking the right equation

This tip is ours, not a rule printed in NCERT. Write down the five quantities: v₀, v, a, t and x. Each equation leaves one out. Spot the quantity the question neither gives nor asks for, then use the equation that doesn't contain it.

The sign mistake that costs marks

Choose a positive direction before you write a single number. Suppose a scooter moving at 10 m s⁻¹ brakes at 2 m s⁻² for 3 s. Take forward as positive. Then a = −2 m s⁻², and

x = 10 × 3 + ½ × (−2) × 3² = 30 − 9 = 21 m.

Drop the minus sign and you get 39 m. Without braking, the scooter would have covered only 30 m, so an answer that says braking carried it further should make you look again.

The reverse habit is just as risky. Negative acceleration does not always mean slowing down. If the velocity is negative too, the object speeds up in the negative direction. The sign of a tells you which way the acceleration points. That's all.

When the equations stop working

Every one of these equations assumes constant acceleration; the Points to Ponder at the end of the chapter say so plainly. A bus pulling out in traffic doesn't accelerate steadily, so the equations won't give its true displacement. The calculus method in Example 2.2 is more general, and the book notes it can handle non-uniform acceleration too.

Try it

A trolley speeds up from 6 m s⁻¹ to 14 m s⁻¹ at a constant 4 m s⁻². How far does it travel? (Answer at the bottom.)

Questions students ask

Answer to the try-it: x = (14² − 6²) ÷ (2 × 4) = (196 − 36) ÷ 8 = 20 m.